Difference between revisions of "2005 AMC 12B Problems/Problem 4"
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| − | + | Trahee's goal was to get an A on <math>80\% \cdot 50 = 40</math> quizzes. She already has A's on <math>22</math> quizzes, so she needs to get A's on <math>40-22=18</math> more. There are <math>50-30=20</math> quizzes left, so she can afford to get less than an A on <math>20-18=\boxed{\text{(B)}2}</math> of them. | |
Here, only the A's matter... No complicated stuff! | Here, only the A's matter... No complicated stuff! | ||
Revision as of 01:59, 18 February 2021
- The following problem is from both the 2005 AMC 12B #4 and 2005 AMC 10B #6, so both problems redirect to this page.
Problem
At the beginning of the school year, Trahee's goal was to earn an A on at least 
of her 
quizzes for the year. She earned an A on 
of the first 
quizzes. If she is to achieve her goal, on at most how many of the remaining quizzes can she earn a grade lower than an A?
Solution
Trahee's goal was to get an A on 
quizzes. She already has A's on quizzes, so she needs to get A's on 
more. There are 
quizzes left, so she can afford to get less than an A on 
of them.
Here, only the A's matter... No complicated stuff!
See also
| 2005 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
| 2005 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 3 |
Followed by Problem 5 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
